Chemistry Matter and Change, Florida Edition

Chapter 2: Data Analysis

Problem of the Week

 
<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::::/sites/dl/free/0078703298/179001/ch02_chapter.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (0.0K)</a>
<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::::/sites/dl/free/0078703298/179001/POWsolutions.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (0.0K)</a>

Teachers Notes:

Problem 1:
No, the mass of each is different. The mass of the gold statue is determined to be 40.3 pounds while that of the bag of sand is 5.2 pounds. The gold statue weighs 7.8 times more than the bag of sand.

1.0 quart x (0.946 liter/quart) x (1dm3/liter) x (1000 cm3/dm3) x (19.3 g Gold/cm3) x (1 pound/453.5 g) = 40.3 pounds

1.0 quart x (0.946 liter/quart) x (1dm3/liter) x (1000 cm3/dm3) x (2.5 g Sand/cm3) x (1 pound/453.5 g) = 5.2 pounds

Problem 2:
The volume of sand that would equal the mass of the gold statue is 7.73 quarts. Almost 8 bags of sand would have to be exchanged for the statue.

40.3 pounds x (453.5 g/1 pound) x (1 cm3/2.5 g) x (1 dm3/1000 cm3) x (1 liter/1 dm3) x 1 quart/ 0.946 liter) = 7.73 quarts.

Problem 3:
Since the density of sand is still less than that of brass, switching the statue with the same volume of sand is not possible. If the statue is hollow, the mass of the statue must be known to determine the volume of sand needed.



Glencoe Online Learning CenterScience HomeProduct InfoSite MapContact Us

The McGraw-Hill CompaniesGlencoe